Hasya Aqilah Binti MOHD Khairi-NUR Amira Shafiqah Binti Zazani- Nurul Syuhada Binti Zainuddin- Group Project ( Analysis OF Fatality RATE PER 100 Milion Vehicles Miles Traveled IN US ) (1) PDF

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MAT523: LINEAR ALGEBRA IITOPIC:ANALYSIS OF FATALITY RATE PER 100 MILION VEHICLES MILES TRAVELED INU USING LEAST SQUARE METHODKUMPULAN: D1CS2492BPREPARED FOR:MOHD RAHIMIE BIN MD NOORPREPARED BY :NAMA STUDENT ID HASYA AQILAH BINTI MOHD KHAIRI 2020828066 NUR AMIRA SHAFIQAH BINTI ZAZANI 2020828394 NURUL...

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MAT523: LINEAR ALGEBRA II TOPIC: ANALYSIS OF FATALITY RATE PER 100 MILION VEHICLES MILES TRAVELED IN U.S USING LEAST SQUARE METHOD KUMPULAN: D1CS2492B PREPARED FOR: MOHD RAHIMIE BIN MD NOOR PREPARED BY : NAMA

STUDENT ID

HASYA AQILAH BINTI MOHD KHAIRI

2020828066

NUR AMIRA SHAFIQAH BINTI ZAZANI

2020828394

NURUL SYUHADA BINTI ZAINUDDIN

2020878544

CONTENT No.

About

Pages

1.

Introduction

1–2

2.

Linear model

3–9

3.

Quadratic model

10 – 17

4.

Cubic model

18 – 25

5.

Conclusion

26

6.

Appendix

27 – 28

1

INTRODUCTION This is a data about the traffic deaths occur in United States over the years (Source: U.S. Department of Transportation, National Highway Traffic Safety Administration). Let focus on the data from the past twenty years (2000 – 2019). From the data, we can see that there is an up and down for both fatalities and fatality rate per 100 million vehicle miles travelled over the years but for fatality rate per 100, 000 registered vehicles, it was decreasing from the year 2000 to 2011 and again there is up and down trends from the year 2012 to 2019. The highest value of fatalities occur in United States is on the year 2002 which the value is 43, 005 and the lowest value is 32, 479 which occur on the year 2011. Next, for the annual percent change of the traffic deaths. From the table, the negative values of annual percent change show that the number of fatalities is decreasing while the positive values show that the number is increasing. We can see that on the year 2008, the annual percent change is -9.7% which show that the number of fatalities is decreasing a lot. On the year 2015, the annual percent change is 8.4% which we can conclude that there is a large increasing number of fatalities on that year. From the large number of fatalities over the years, it was divided into two categories which are fatality rate per 100 million vehicle miles travelled and fatalities rate per 100, 000 registered vehicles. All these three categories are related to the annual percent change of the traffic death and the year. But for this mini project assignment, we decide to make a comparing between the annual percent change and the fatality rate per 100 million vehicle miles travelled. The annual percent change will take the role as the x-variable and the fatality rate per 100 million vehicle miles travelled will take the role as the y-variable. The data we take is as below:

2

Annual percent change (x) 0.5 0.6 1.9 -0.3 -0.1 1.6 -1.8 -3.4 -9.3 -9.7 -2.6 -1.6 4.0 -2.6 -0.5 8.4 6.5 -0.9 -1.7 -2.0

Fatality rate per 100 million vehicle miles travelled (y) 1.53 1.51 1.51 1.48 1.44 1.46 1.42 1.36 1.26 1.13 1.11 1.10 1.14 1.10 1.08 1.15 1.19 1.17 1.14 1.10

3

Table of data: Number of data x y

Number of Data x y

1

2

3

4

5

6

7

8

9

10

0.5 1.53

0.6 1.51

1.9 1.51

-0.3 1.48

-0.1 1.44

1.6 1.46

-1.8 1.42

-3.4 1.36

-9.3 1.26

-9.7 1.13

11

12

13

14

15

16

17

18

19

20

-2.6 1.11

-1.6 1.1

4 1.14

-2.6 1.1

-0.5 1.08

8.4 1.15

6.5 1.19

-0.9 1.17

-1.7 1.14

-2 1.1

Linear Model

a) Find the equation of the curve by list square method.

𝒚 = 𝒂 + 𝒄𝒙

(0.5,1.53) → (0.6,1.51) → (1.9,1.51) → (−0.3,1.48) → (−0.1,1.44) → (1.6,1.46) → (−1.8,1.42) → (−3.4,1.36) → (−9.3,1.26) → (−9.7,1.13) → (−2.6,1.11) → (−1.6,1.1) → (4,1.14) → (−2.6,1.1) → (−0.5,1.08) → (8.4,1.15) → (6.5,1.19) → (−0.9,1.17) → (−1.7,1.14) → (−2,1.1) →

1.53 = 0.5𝑚 + 𝑐 1.5 = 0.6𝑚 + 𝑐 1.51 = 1.9𝑚 + 𝑐 1.48 = −0.3𝑚 + 𝑐 1.44 = −0.1𝑚 + 𝑐 1.46 = 1.6𝑚 + 𝑐 1.42 = −1,8𝑚 + 𝑐 1.36 = −3.4𝑚 + 𝑐 1.26 = −9.3𝑚 + 𝑐 1.13 = −9.7𝑚 + 𝑐 1.11 = −2.6𝑚 + 𝑐 1.1 = −1.6𝑚 + 𝑐 1.14 = 4𝑚 + 𝑐 1.1 = −2.6𝑚 + 𝑐 1.0 = −0.5𝑚 + 𝑐 1.1 = 8.4𝑚 + 𝑐 1.19 = 6.5𝑚 + 𝑐 1.17 = −0.9𝑚 + 𝑐 1.14 = −1.7𝑚 + 𝑐 1.1 = −2𝑚 + 𝑐

4

𝐴𝑇

=[

1 1 1 1 1 1 1 1 1 1 𝐴= 1 1 1 1 1 1 1 1 1 [1

0.5 0.6 1.9 −0.3 −0.1 1.6 −1.8 −3.4 −9.3 𝑎 −9.7 𝑥=( ) 𝑏 −2.6 −1.6 4 −2.6 −0.5 8.4 6.5 −0.9 −1.7 −2 ]

1.53 1.51 1.51 1.48 1.44 1.46 1.42 1.36 1.26 1.13 𝑏= 1.11 1.1 1.14 1.1 1.08 1.15 1.19 1.17 1.14 [ 1.1 ]

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 ] 0.5 0.6 1.9 −0.3 −0.1 1.6 −1.8 −3.4 −9.3 −9.7 −2.6 −1.6 4 −2.6 −0.5 8.4 6.5 −0.9 −1.7 −2

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 ] 1 𝐴𝑇 𝐴 = [ 0.5 0.6 1.9 −0.3 −0.1 1.6 −1.8 −3.4 −9.3 −9.7 −2.6 −1.6 4 −2.6 −0.5 8.4 6.5 −0.9 −1.7 −2

𝐴𝑇 𝐴 = [

20 −13 ] −13 355.1

1

1

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 [1

0.5 0.6 1.9 −0.3 −0.1 1.6 −1.8 −3.4 −9.3 −9.7 −2.6 −1.6 4 −2.6 −0.5 8.4 6.5 −0.9 −1.7 −2 ]

5

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 𝐴𝑇 𝑏 = [ ] 0.5 0.6 1.9 −0.3 −0.1 1.6 −1.8 −3.4 −9.3 −9.7 −2.6 −1.6 4 −2.6 −0.5 8.4 6.5 −0.9 −1.7 −2

𝐴𝑇 𝑏 = [

25.38 ] −14.85

𝐴𝑇 𝐴 = [

20 −13 ] −13 355.1

[

𝐴𝑇 𝑏 = [ 𝐴𝑥 = 𝑏 𝑨𝑻 𝑨𝒙 = 𝑨𝑻 𝒃

25.38 20 −13 𝑎 ][ ] = [ ] 𝑏 −14.85 −13 355.1

13 1.269 ] 𝑅2→𝑅2+13𝑅1 [ 1 − 20| → −14.85 −13 355.1 13 13 1.269 𝑅1→𝑅1+ 20𝑅2 1 0 1.27 1 − 20| [ ]→ [ | ] 0 1 0.01 0 1 0.0050 20 −13 25.38 ]→ | [ −13 355.1 −14.85

𝑅1→

So,

1 𝑅1 20

𝑎 = 1.27

Thus, the equation of the curve is

𝑦 = 𝑎 + 𝑏𝑥

𝑏 = 0.01

𝑦 = 1.27 + 0.01𝑥

1

1

25.38 ] −14.85

1.53 1.51 1.51 1.48 1.44 1.46 1.42 1.36 1.26 1.13 1.11 1.1 1.14 1.1 1.08 1.15 1.19 1.17 1.14 [ 1.1 ]

6 b) Find the error vector and the magnitude of this error vector. NUMBER OF DATA x

1 0.56

2 0.6

3 1.9

4 -0.3

5 -0.1

Real y Approximate y

1.53 1.28

1.51 1.28

1.51 1.29

1.48 1.27

1.44 1.27

NUMBER OF DATA

6

7

8

9

10

x

1.6

-1.8

-3.4

-9.3

-9.7

Real y

1.46

1.42

1.36

1.26

1.13

Approximate y

1.29

1.25

1.24

1.17

1.18

NUMBER OF DATA

11

12

13

14

15

x

-2.6

-1.6

4

-2.6

-0.5

Real y

1.11

1.1

1.14

1.1

1.08

Approximate y

1.24

1.25

1.31

1.24

1.27

NUMBER OF DATA

16

17

18

19

20

x Real y

8.4 1.15

6.5 1.19

-0.9 1.17

-1.7 1.14

-2 1.1

Approximate y

1.35

1.34

1.26

1.25

1.25

Error= real data – approximate data

e1=1.53-1.28= 0.25 e2=1.51-1.28= 0.23 e3 =1.51-1.29= 0.22 e4 =1.48-1.27= 0.21 e5 =1.44-1.27= 0.17 e6 =1.46-1.29= 0.17 e7=1.42-1.25= 0.17 e8=1.36-1.24= 0.12 e9 =1.26-1.17= 0.09 e10=1.131.18= -0.05

e11=1.11-1.24= -0.13 e12=1.10-1.25= -0.15 e13=1.14-1.31= -0.17 e14=1.1-1.24= -0.14 e15=1.08-1.27= -0.19 e16=1.15-1.35= -0.16 e17=1.19-1.34= -0.15 e18=1.17-1.26= -0.09 e19=1.14-1.25= -0.11 e20=1.1-1.25= -0.15

7

|e|=√𝑒1 2 + 𝑒2 2 + 𝑒3 2 +. . . +𝑒20 2

|e|= √

(0.25)2 + (0.23)2 + (0.22)2 + (0.21)2 + (0.17)2 + (0.17)2(0.17)2 + (0.12)2 + (0.09)2 + (−0.05)2 + (−0.15)2 + (−0.17)2 + (−0.14)2 + (−0.19)2 + (−0.16)2 + (−0.15)2 + (−0.09)2 + (−0.11)2 + (−0.15)2

+(−0.13)2

|e|=√0.5344 |e|=0.731

8

c) Plot the graphs: i. Points of scattered data and best fit curve that you have calculated. ii. Graph of residuals (error vector) around the x-axis (y = 0) x 0.5 0.6 1.9 -0.3 -0.1 1.6 -1.8 -3.4 -9.3 -9.7 -2.6 -1.6 4 -2.6 -0.5 8.4 6.5 -0.9 -1.7 -2

Real y 1.53 1.51 1.51 1.48 1.44 1.46 1.42 1.36 1.26 1.13 1.11 1.1 1.14 1.1 1.08 1.15 1.19 1.17 1.14 1.1

y-approximate 1.28 1.28 1.29 1.27 1.27 1.29 1.25 1.24 1.17 1.18 1.24 1.25 1.31 1.24 1.27 1.35 1.34 1.26 1.25 1.25

Error 0.25 0.23 0.22 0.21 0.17 0.17 0.17 0.12 0.09 -0.05 -0.13 -0.15 -0.17 -0.14 -0.19 -0.16 -0.15 -0.09 -0.11 -0.15

9

Fatality rate per 100 million vehicles miles traveled

Fatality rate per 100 million vehicles miles traveled in US over Annual Percent Change 1.8 1.6

1.4 1.2 1 0.8 0.6 0.4 0.2 0 -12

-10

-8

-6

-4

-2

0

2

4

6

8

10

Annual percent change

Fatality rate per 100 million vehicles miles traveled

Fatality rate per 100 million vehicles miles traveled in US over Annual Percent Change 1.8 1.6 1.4 1.2 1 0.8 0.6 0.4 0.2

-12

-10

-8

-6

-4

0 -2 0 2 Annual percent change

4

6

8

10

10 Quadratic Model

a) Equation of the curve by using the least square method. 1.53 = 𝑎 + 0.5𝑏 + 0.25𝑐

1.51 = 𝑎 + 0.6𝑏 + 0.36𝑐

𝒚 = 𝒂 + 𝒃𝒙 + 𝒄𝒙𝟐

1.11 = 𝑎 + (−2.6)𝑏 + 6. 76𝑐 1.1 = 𝑎 + (−1.6)𝑏 + 2.56𝑐

1.51 = 𝑎 + 1.9𝑏 + 3.61𝑐

1.14 = 𝑎 + 4𝑏 + 16𝑐

1.48 = 𝑎 + (−0.3)𝑏 + 0. 09𝑐

1.1 = 𝑎 + (−2.6)𝑏 + 6.76𝑐

1.44 = 𝑎 + (−0.1)𝑏 + 0. 01𝑐

1.08 = 𝑎 + (−0.5)𝑏 + 0.25𝑐

1.46 = 𝑎 + 1.6𝑏 + 2.56𝑐

1.15 = 𝑎 + 8.4𝑏 + 70.56𝑐

1.42 = 𝑎 + (−1.8)𝑏 + 3.24𝑐

1.19 = 𝑎 + 6.5𝑏 + 42.25𝑐

1.13 = 𝑎 + (−9.7)𝑏 + 94.09𝑐

1.1 = 𝑎 + (−2)𝑏 + 4𝑐

1.36 = 𝑎 + (−3.4)𝑏 + 11.56𝑐

1.17 = 𝑎 + (−0.9)𝑏 + 0.81𝑐

1.26 = 𝑎 + (−9.3)𝑏 + 86.49𝑐

1.14 = 𝑎 + (−1.7)𝑏 + 2.89𝑐

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 [1

0.5 0.6 1.9 −0.3 −0.1 1.6 −1.8 −3.4 −9.3 −9.7 −2.6 −1.6 4 −2.6 −0.5 8.4 6.5 −0.9 −1.7 −2

Ax=b

0.25 0.36 3.61 0.09 0.01 2.5 3.24 11.56 86.49 𝑎 94.09 𝑏 ] = [ 6.76 𝑐 2.56 16 6.76 0.25 70.56 42.25 0.81 2.89 4 ]

1.53 1.51 1.51 1.48 1.44 1.46 1.42 1.36 1.26 1.13 1.11 1.1 1.14 1.1 1.08 1.15 1.19 1.17 1.14 [ 1.1 ]

11

1 1 1 1 1 1 1 1 1 1 𝐴= 1 1 1 1 1 1 1 1 1 [1

0.5 0.6 1.9 −0.3 −0.1 1.6 −1.8 −3.4 −9.3 −9.7 −2.6 −1.6 4 −2.6 −0.5 8.4 6.5 −0.9 −1.7 −2

1 1 1 1 𝐴𝑇 = [ 0.5 0.6 1.9 −0.3 0.25 0.36 3.61 0.09

1 𝐴𝑇 𝐴 = [ 0.5 0.25

1 1 1.9 0.6 0.36 3.61

0.25 0.36 3.61 0.09 0.01 2.5 3.24 11.56 86.49 94.09 6.76 2.56 16 6.76 0.25 70.56 42.25 0.81 2.89 4 ]

1 1 1 −0.1 1.6 −1.8 0.01 2.5 3.24

1 1 1 −0.3 −0.1 1.6 0.09 0.01 2.5

1 −1.8 3.24

𝑎 𝑏 𝑥=[ ] 𝑐 𝑑

1 1 1 1 1 1 −3.4 −9.3 −9.7 −2.6 −1.6 4 11.56 86.49 94.09 6.76 2.56 16

1 1 1 −3.4 −9.3 −9.7 11.56 86.49 94.09

1 1 1 1 1 1 −2.6 −1.6 4 −2.6 −0.5 8.4 6.76 2.56 16 6.76 0.25 70.56

20 −13 355.04 𝐴𝑇 𝐴 = [ −13 355.1 −872.68 ] 355.04 −872.68 23639.8574

1 1 1 1 1 1 1 1 1 1 1 1 1 1 0.6 1.9 −0.3 −0.1 1.6 −1.8 −3.4 −9.3 −9.7 −2.6 −1.6 4 −2.6 0.25 0.36 3.61 0.09 0.01 2.5 3.24 11.56 86.49 94.09 6.76 2.56 16 6.76

𝐴𝑇 𝑏 = [ 0.5

1 1 −2.6 −0.5 6.76 0.25

1.53 1.51 1.51 1.48 1.44 1.46 1.42 1.36 1.26 𝑏 = 1.13 1.11 1.1 1.14 1.1 1.08 1.15 1.19 1.17 1.14 [ 1.1 ]

1 8.4 70.56

1 1 −0.9 6.5 42.25 0.81

1 1 1 1 6.5 −0.9 −1.7 −2 ] 4 42.25 0.81 2.89

1 1 1 1 1 1 1 1 1 1 1 −1.7 −2] 1 1 2.89 4 1 1 1 1 1 1 1 1 [1

0.5 0.6 1.9 −0.3 −0.1 1.6 −1.8 −3.4 −9.3 −9.7 −2.6 −1.6 4 −2.6 −0.5 8.4 6.5 −0.9 −1.7 −2

0.25 0.36 3.61 0.09 0.01 2.5 3.24 11.56 86.49 94.09 6.76 2.56 16 6.76 0.25 70.56 42.25 0.81 2.89 4 ]

1.53 1.51 1.51 1.48 1.44 1.46 1.42 1.36 1.26 1 1 1 1 1 1 6.5 −0.9 −1.7 −2] 1.13 −0.5 8.4 0.25 70.56 42.25 0.81 2.89 4 1.11 1.1 1.14 1.1 1.08 1.15 1.19 1.17 1.14 [ 1.1 ]

12

20 𝐴𝑇 𝐴 = [ −13

−13 355.1

𝐴𝑇 𝑏 = [

355.04 −872 .68

355.04 −872 .68 23639.8574 20 [ −13 355.04

]

25.38 −14 .853

422.1257

𝑨𝑻 𝑨𝒙 = 𝑨𝑻 𝒃

]

𝐴𝑇 𝑏

−14.853 ] 25.38 = [ 422.1257

𝑎 355.04 25.38 −13 355.1 −872.68 ] [ 𝑏 ] = [ −14.853 ] −872.68 23639.8574 𝑐 422.1257

−13 20 [ −13 355.1 355.04 −872.68

25.38 355.04 | −14.853 ] −872.68 23639.8574 422.1257

1 −13 25.38 20 355.04 𝑅 ⟶𝑅1 20 1 [ −13 355.1 −872.68 | −14.853 ] → 355.04 −872.68 23639.8574 422.1257

1.269 −0.65 17.752 1 𝑅2 +13𝑅1 ⟶𝑅2 [ −13 355.1 −872.68 | −14.853 ] → 355.04 −872.68 23639.8574 422.1257 [

17.752 1 1.269 −0.65 𝑅3 −355.04𝑅1 ⟶𝑅3 | 0 1.644 ] → 346.65 −641.904 355.04 −872.68 23639.8574 422.1257

1 [0 0

1 [0 0 1 [0 0

1 [0 0

1 −0.65 17.752 1.269 𝑅 ⟶𝑅2 346.65 2 1.644 ] → 346.65 −641.904 | −641.904 17337.18732 −28.42006

1.269 −0.65 17.752 𝑅3 +641.904𝑅2 ⟶𝑅3 ] → 1 −1.85173518 | 4.742535699 × 10−3 −28.42006 −641.904 17337.18732

−0.65 1 0

−0.65 1 0

1 1.269 17.752 𝑅 ⟶𝑅3 16148.5511 3 −3 −1.85173518 | 4.742535699 × 10 ] → 16148.5511 −25.37580736

1.269 17.752 −1.85173518 |4.742535699 × 10−3] 1 −1.5713984 × 10−3

13

𝑎 − 0.65𝑏 + 17 .752𝑐 = 1.269 𝑏 − 1.85173518𝑐 = 4.742535699 × 10−3 𝑐 = −1.5713984 × 10−3

𝑏 − 1.85173518(−1.5713984 × 10−3 ) = 4.742535699 × 10−3

𝑏 = 1.832722 × 10−3

𝑎 − 0.65(1.832722 × 10−3 ) + 17.752(−1.5713984 × 10−3 ) = 1.269 𝑎 = 1.298086734 So,

𝑎 = 1.298086734,

The equation is,

𝑏 = 1.832722 × 10−3 ,

𝑐 = −1.5713984 × 10−3

𝑦 = 1.298086734 + 1.832722 × 10−3 𝑥 − 1.5713984 × 10−3 𝑥 2 Or

y = -0.0016x2 + 0.0018x + 1.2981

14

b) Find the error vector and the magnitude of this error vector. Number of data 1 2 3 4 5 0.5 0.6 1.9 -0.3 -0.1 x 1.53 1.51 1.51 1.48 1.44 real y approximate y 1.29861 1.29862 1.2959 1.2974 1.29789

Number of data 6 7 x 1.6 -1.8 real y 1.46 1.42 approximate y 1.297 1.2897

8 9 10 -3.4 -9.3 -9.7 1.36 1.26 1.13 1.27369 1.14513 1.13246

Number of 11 12 13 14 data -2.6 -1.6 4 -2.6 x 1.11 1.1 1.14 1.1 real y approximate 1.2827 1.29113 1.28028 1.2827 y

Number of data 16 17 18 19 x 8.4 6.5 -0.9 -1.7 Real y 1.15 1.19 1.17 1.14 approximate y 1.2026 1.24361 1.29516 1.29043 Error=real data – Approximated data 𝑒1 = 1.53 − 1.29861 = 0.23139

𝑒2 = 1.51 − 1.29862 = 0.21138 𝑒3 = 1.51 − 1.2959 = 0. 2141

𝑒4 = 1.48 − 1.2974 = 0. 1826

𝑒5 = 1.44 − 1.29789 = 0.14211

20 -2 1.1 1.28814

𝑒2 2 = 0.04468

𝑒3 2 = 0.04583

𝑒4 2 = 0.03334

𝑒5 2 = 0.02020

𝑒9 = 1.26 − 1.14513 = 0.11487

𝑒9 2 = 0.01320

𝑒10 = 1.13 − 1.13246 = −0.00246

1.29678

𝑒1 2 = 0.05354

𝑒6 2 =0.02656

𝑒8 = 1.36 − 1.27369 = 0.08631

-0.5 1.08

|𝑒| = √𝑒1 2 + 𝑒2 2 + 𝑒3 2 + ⋯ 𝑒20 2

𝑒6 = 1.46 − 1.297 = 0.163

𝑒7 = 1.42 − 1.2897 = 0. 1303

15

𝑒7 2 = 0.01698

𝑒8 2 = 0.00745

𝑒10 2 = 6.0516 × 10−6

15

𝑒11 = 1.11 − 1.2827 = −0.1727 𝑒12 = 1.1 − 1.29113 = −0.19113

𝑒13 = 1.14 − 1.28028 = −0.14028

𝑒11 2 = 0.02983 𝑒12 2 = 0.03653

𝑒13 2 = 0.01968

𝑒14 = 1.1 − 1.2827 = −0.1827

𝑒14 2 = 0.03338

𝑒17 = 1.19 − 1.24361 = −0.05361

𝑒17 2 = 2.87403 × 10−3

𝑒15 = 1.08 − 1.29678 = −0.21678 𝑒16 = 1.15 − 1.2026 = −0.0526

𝑒18 = 1.17 − 1.29516 = −0.12516

𝑒19 = 1.14 − 1.29043 = −0.15043

𝑒20 = 1.1 − 1. 28814 = −0.18814 |𝑒| = √𝑒1 2 + 𝑒2 2 + 𝑒3 2 + ⋯ 𝑒20 2

𝑒15 2 = 0.04699

𝑒16 2 = 2.76676 × 10−3 𝑒18 2 = 0.01567

𝑒19 2 = 0.02263

𝑒20 2 = 0.03540

∑ 𝑒𝑛 2 = 0.05354 + 0.04468 + 0.04583 + 0.03334 + 0.02020 + 0.02656 + 0.0169

+ 0.00745 + 0.01320 + 6.0516 × 10−6 + 0.02983 + 0. 03653 + 0.01968 + 0.03338 + 0.04699 + 2.76676 × 10−3 + 2.87403 × 10 −3 + 0.01567 + 0.02263 + 0.03540

e = √0.5075368416 e = 0.7124161997

16

c) Plot the graphs: i. Points of scattered data and best fit curve that you have calculated. ii. Graph of residuals (error vector) around the x-axis (y = 0)

x 0.5 0.6 1.9 -0.3 -0.1 1.6 -1.8 -3.4 -9.3 -9.7 -2.6 -1.6 4 -2.6 -0.5 8.4 6.5 -0.9 -1.7 -2

y 1.53 1.51 1.51 1.48 1.44 1.46 1.42 1.36 1.26 1.13 1.11 1.1 1.14 1.1 1.08 1.15 1.19 1.17 1.14 1.1

y-approximate 1.29861 1.29862 1.2959 1.2974 1.29789 1.297 1.2897 1.27369 1.14513 1.13246 1.2827 1.29113 1.28028 1.2827 1.29678 1.2026 1.24361 1.29516 1.29043 1.28814

Error 0.23139 0.21138 0.2141 0.1826 0.14211 0.163 0.1303 0.08631 0.11487 -0.00246 -0.1727 -0.19113 -0.14028 -0.1827 -0.21678 -0.0526 -0.05361 -0.12516 -0.15043 -0.18814

17

Fatality rate 100 milion vehicles miles traveled in US over Annual Percent Change Fatality rate per 100 million vehicles miles traveled

1.8 1.6 1.4 1.2 1 0.8 0.6

0.4 0.2 0

-12

-10

-8

-6

-4

-2

0

2

4

6

8

10

8

10

Annual percent change

Fatality rate per 100 million vehicles miles traveled in US over Annual Percent Change Fatality rate per 100 million vehicles miles traveled

1.8

-12

1.6 1.4 1.2 1 0.8 0.6 0.4 0.2 0 -10

-8

-6

-4

-2

0

Annual percent change

2

4

6

18 Cubic Model

a) Equation of the curve by using the least square method.

𝒚 = 𝒂 + 𝒃𝒙 + 𝒄𝒙𝟐 + 𝒅𝒙𝟑

1.53 = 𝑎 + 𝑏 (0.5) + 𝑐(0.5)2 + 𝑑 (0.5)3

→ (0.5, 1.53)

1.48 = 𝑎 + 𝑏 (−0.3) + 𝑐(−0.3)2 + 𝑑 (−0.3)3

→ (−0.3, 1.48)

1.42 = 𝑎 + 𝑏 (−1.8) + 𝑐(−1.8)2 + 𝑑 (−1.8)3

→ (−1.8, 1.42)

1.51 = 𝑎 + 𝑏 (0.6) + 𝑐(0.6)2 + 𝑑 (0.6)3

1.51 = 𝑎 + 𝑏 (1.9) + 𝑐(1.9)2 + 𝑑 (1.9)3

1.44 = 𝑎 + 𝑏 (−0.1) + 𝑐(−0.1)2 + 𝑑 (−0.1)3 1.46 = 𝑎 + 𝑏 (1.6) + 𝑐(1.6)2 + 𝑑 (1.6)3

1.36 = 𝑎 + 𝑏 (−3.4) + 𝑐(−3.4)2 + 𝑑 (−3.4)3

1.26 = 𝑎 + 𝑏 (−9.3) + 𝑐(−9.3)2 + 𝑑 (−9.3)3

1.13 = 𝑎 + 𝑏 (−9.7) + 𝑐(−9.7)2 + 𝑑 (−9.7)3

1.10 = 𝑎 + 𝑏 (−1.6) + 𝑐...